I am asked to solve the following differential equation:

$$ x y^2 y' = x+1 $$

My process was

$$ \begin{align*} x y^2 y' &= x+1\\ xy^2 \frac{dy}{dx} &= x+1\\ y^2 dy &= \frac{x+1}{x} dx\\ \int y^2 dy &= \int \frac{x+1}{x} dx\\ \int y^2 dy &= \int dx + \int \frac{1}{x} dx\\ \frac{y^3}{3} &= x + \ln |x| + C\\ y &= \sqrt[3]{3 \left( x + \ln |x| + C \right)} \end{align*} $$

but when I was checking my result on Wolfram I noticed that it was given in a different way.

enter image description here

Is my result incorrect? What caused the results to be different? Is it the absolute value sign of the $\ln$?

Thank you.

  • $\begingroup$ Looks the same to me. $\endgroup$
    – A.Γ.
    Aug 12, 2016 at 21:15
  • 1
    $\begingroup$ Well, that is simply because there are three cubic roots of most numbers. If $y=y_1$ is a solution, then $y=\exp\left(\frac{2\pi\text{i}}{3}\right)y_1$ and $y=\exp\left(-\frac{2\pi\text{i}}{3}\right)y_1$ are also solutions. Only when you put a restriction that the solution be real-valued, then you will get only one of them. For Wolfram Alpha, $(-1)^{1/3}=\exp\left(\frac{\pi\text{i}}{3}\right)$. $\endgroup$ Aug 12, 2016 at 21:16
  • $\begingroup$ Your answer is correct. Wolfram Alpha just simplified it. $\endgroup$
    – user322313
    Aug 12, 2016 at 21:17

1 Answer 1


So, every real number has three cube roots: a real one, and two complex ones. The cube roots of 1, for instance, are $1, -\frac{1}{2}+\frac{\sqrt{3}}{2}i$, and $-\frac{1}{2}-\frac{\sqrt{3}}{2}i$. When you took the cube root in solving for $y$, you only considered the real cube root (which is totally fine if you're only interested in real-values functions); WolframAlpha was more general, and gave solutions for all three cube roots. Sort of like how when you have $y^2 = x$, this gives two solutions for $y$: $y = \sqrt{x}$ and $y = -\sqrt{x}$.

Also, with regards to the absolute value, both $\ln(x)$ and $\ln(|x|)$ are valid antiderivatives of $\frac{1}{x}$, so you're correct there.

  • 1
    $\begingroup$ Thank you @florence, best regards! $\endgroup$
    – bru1987
    Aug 12, 2016 at 22:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.